Akif Imtiaz

Exercise Set 1.6 45

Since the proof of Theorem 1.4.8 (b) was omitted, we note that

Because Ak is invertible, Theorem 1.6.4 allows us to conclude that Akx = 0 has only the
trivial solution.

25. Suppose that x1 is a fixed matrix which satisfies the equation Ax1 = b. Further, let x be any
matrix whatsoever which satisfies the equation Ax = b. We must then show that there is a
matrix x0 which satisfies both of the equations x = x1 + x0 and Ax0 = 0.
Clearly, the first equation implies that
x0 = x – x1

This candidate for x0 will satisfy the second equation because
Ax0 = A(x – x1) = Ax – Ax1 = b – b = 0

We must also show that if both Ax1 = b and Ax0 = 0, then A(x1 + x0) = b. But
A(x1 + x0) = Ax1 + Ax0 = b + 0 = b

27. (a) x ≠ 0 and x ≠ y

(b) x ≠ 0 and y ≠ 0

(c) x ≠ y and x ≠ –y

Gaussian elimination has to be performed on (A  I) to find A–1. Then the product
A–1B is performed, to find x. Instead, use Gaussian elimination on (A  B) to find x. There
are fewer steps in the Gaussian elimination, since (A  B) is a m × (n+1) matrix in general,
or n × (n+1) where A is square (n × n). Compare this with (A  I) which is n × (2n) in the
inversion approach. Also, the inversion approach only works for A n × n and invertible.

46 Exercise Set 1.6

29. No. The system of equations Ax = x is equivalent to the system (A – I)x = 0. For this
system to have a unique solution, A – I must be invertible. If, for instance, A = I, then
any vector x will be a solution to the system of equations Ax = x.

Note that if x ≠ 0 is a solution to the equation Ax = x, then so is kx for any real number k.
A unique solution can only exist if A – I is invertible, in which case, x = 0.

31. Let A and B be square matrices of the same size. If either A or B is singular, then AB is
singular.

EXERCISE SET 1.7

7. The matrix A fails to be invertible if and only if a + b – 1 = 0 and the matrix B fails to be
invertible if and only if 2a – 3b – 7 = 0. For both of these conditions to hold, we must have
a = 2 and b = –1.

9. We know that A and B will commute if and only if

AB = 2 1 a b  =  2a + b 2b + d 
   d   
 1 −5   b   a − 5b b − 5d 

is symmetric. So 2b + d = a – 5b, from which it follows that a – d = 7b.

11. (b) Clearly

 ka11 ka12 ka13  3 k 0 0
 ka21 ka22 ka23   5k 
A =    0 0 

 ka31 ka32 ka33   0 0 7 k 

for any real number k = 0.

47

48 Exercise Set 1.7

13. We verify the result for the matrix A by finding its inverse.

 −1 2 5 1 0 0 

 0 1 3 0 1 0
 
 0 0 −4 0 0 1 

 1 −2 −5 −1 0 0 

0 1 3 0 1 0 Multiply Row 1 by –1
  and Row 3 by –1/4.
 0 0 1 0 0 −1 4 

 1 0 1 −1 2 0 Add 2 times Row 2 to
  Row 1 and –3 times
0 1 0 0 1 34 Row 3 to Row 2.
 
 0 0 1 0 0 −1 4  Add –1 times Row
3 to Row 1.
 1 0 0 −1 2 14
 
0 1 0 0 1 3 4
 
 0 0 1 0 0 −1 4 

Thus A–1 is indeed upper triangular.

15. (a) If A is symmetric, then AT = A. Then (A2)T = (AA)T = ATAT = A . A = A2, so A2
is symmetric.

(b) We have from part (a) that

(2A2 – 3A + I)T = 2(A2)T – 3AT + IT = 2A2 – 3A + I

17. From Theorem 1.7.1(b), we have if A is an n × n upper triangular matrix, so is A2. By
induction, if A is an n × n upper triangular matrix, so is Ak, k = 1, 2, 3, . . . We note that the
identity matrix In = A0 is also upper triangular. Next, if A is n × n upper triangular, and K
is any (real) scalar, then KA is upper triangular. Also, if A and B are n × n upper triangular
matrices, then so is A+B. These facts allow us to conclude if p(x) is any (real) polynomial,
and A is n × n upper triangular, then P(A) is an n × n upper triangular matrix.

Exercise Set 1.7 49

19. Let

x 0 0
 
A =  0 y 0 

 0 0 z 

Then if A2 – 3A – 4I = O, we have

 x2 0 0   x 0 0 1 0 0
 y2 0   0 y   
0 0 z2  − 3  0 0 0  − 4  0 1 0  =O
 
 0   z   0 0 1 

This leads to the system of equations
x2 – 3x – 4 = 0
y2 – 3y – 4 = 0
z2 – 3z – 4 = 0

which has the solutions x = 4, –1, y = 4, –1, z = 4, –1. Hence, there are 8 possible choices
for x, y, and z, respectively, namely (4, 4, 4), (4, 4, –1), (4, –1, 4), (4, –1, –1), (–1, 4, 4),
(–1, 4, –1), (–1, –1, 4), and (–1, –1, –1).

23. The matrix

A =  0 1
 −1 
 0 

is skew-symmetric but

AA = A2 =  −1 0
 0 
 −1 

is not skew-symmetric. Therefore, the result does not hold.

In general, suppose that A and B are commuting skew-symmetric matrices. Then
(AB)T = (BA)T = AT BT = (–A)(–B) = AB, so that AB is symmetric rather than skew-
symmetric. [We note that if A and B are skew-symmetric and their product is symmetric,
then AB = (AB)T = BT AT = (–B)(–A) = BA, so the matrices commute and thus skew-
symmetric matrices, too, commute if and only if their product is symmetric.]

50 Exercise Set 1.7

25. Let

A =  x y
 0 
 z 

Then

 x3 ( )y 
A3 =  x2 + xz + z2  =  1 30 
z3   0 
 0   −8 


Hence, x3 = 1 which implies that x = 1, and z3 = –8 which implies that z = –2. Therefore,
3y = 30 and thus y = 10.

27. To multiply two diagonal matrices, multiply their corresponding diagonal elements to obtain
a new diagonal matrix. Thus, if D1 and D2 are diagonal matrices with diagonal elements
d1, . . ., dn and e1, . . ., en respectively, then D1D2 is a diagonal matrix with diagonal elements
d1e1, . . ., dnen. The proof follows directly from the definition of matrix multiplication.

29. In general, let A = [aij]n × n denote a lower triangular matrix with no zeros on or below the
diagonal and let Ax = b denote the system of equations where b = [b1, b2, . . ., bn]T. Since A
is lower triangular, the first row of A yields the equation a11x1 = b1. Since a11 ≠ 0, we can
solve for x1. Next, the second row of A yields the equation a21x1 + a22x2 = b2. Since we
know x1 and since a22 ≠ 0, we can solve for x2. Continuing in this way, we can solve for
successive values of xi by back substituting all of the previously found values x1, x2, . . ., xi–1.

SUPPLEMENTARY EXERCISES 1

1.

3 −4 x 
 5 
 5 

4 3 y 
 5 5 

 1 −4 5 x 
 3 3 
 
4 3  Multiply Row 1 by 5/3.
 5 5 y 
Add –4/5 times Row 1
 1 −4 5  to Row 2.
 3 x 
  Multiply Row 2 by 3/5.
3
Add –4/3 times Row 2
 0 5 − 4 x + y  to Row 1.
 3 3 

 1 −4 5 
 3 x 
 
3

 0 1 − 4 x + 3 y 
 5 5 

 1 0 3 x + 4 y 
 5 5 
 
 4 3 
 0 1 − 5 x + 5 y 

51

52 Supplementary Exercises 1

Thus, x′ = 3–x + 4– y
55

y′ = – 4–x + 3–y
55

3. We denote the system of equations by
a11x1 + a12x2 + a13x3 + a14x4 = 0
a21x1 + a22x2 + a23x3 + a24x4 = 0

If we substitute both sets of values for x1, x2, x3, and x4 into the first equation, we obtain
a11 – a12 + a13 + 2a14 = 0

2a11 + 3a13 – 2a14 = 0

where a11, a12, a13, and a14 are variables. If we substitute both sets of values for x1, x2, x3,
and x4 into the second equation, we obtain

a21 – a22 + a23 + 2a24 = 0
2a21 + 3a23 – a24 = 0

where a21, a22, a23, and a24 are again variables. The two systems above both yield the matrix

 1 −1 1 2 0 
 
 2 0 3 −1 0 

which reduces to

 1 0 3 2 −1 2 0 
 
 0 1 12 −5 2 0 

This implies that

a11 = –(3/2)a13 + (1/2)a14
a12 = –(1/2)a13 + (5/2)a14

Supplementary Exercises 1 53

and similarly,

a21 = (–3/2)a23 + (1/2)a24
a22 = (–1/2)a23 + (5/2)a24

As long as our choice of values for the numbers aij is consistent with the above, then
the system will have a solution. For simplicity, and to insure that neither equation is a
multiple of the other, we let a13 = a14 = –1 and a23 = 0, a24 = 2. This means that
a11 = 1, a12 = –2, a21 = 1, and a22 = 5, so that the system becomes

x1 – 2x2 – x3 – x4 = 0

x1 + 5x2 + 2x4 = 0

Of course, this is just one of infinitely many possibilities.

5. As in Exercise 4, we reduce the system to the equations

1+ 5z
x=

4
35 – 9z
y=

4

Since x, y, and z must all be positive integers, we have z > 0 and 35 – 9z > 0 or 4 > z. Thus
we need only check the three values z = 1, 2, 3 to see whether or not they produce integer
solutions for x and y. This yields the unique solution x = 4, y = 2, z = 3.

9. Note that K must be a 2 × 2 matrix. Let

K = a b
 
 c d 

Then

 1 4  a b2 0 0   8 6 −6 
   c 1 −1   
 −2 3   d   0  =  6 −1 1 
−2     −4 0
 1 0 

54 Supplementary Exercises 1

or

 1 4  2a b −b   8 6 −6 
   2c d −d   
 −2 3    =  6 −1 1 
−2   −4 0
 1 0 

or

 2a + 8c b + 4d −b − 4d   8 6 −6 
 −2b + 3d   
 −4a + 6c 2b − 3d  =  6 −1 1 
b − 2d
 2a − 4c −b + 2d   −4 0 0 

Thus

2a + 8c = 8
b + 4d = 6

– 4a + 6c = 6
– 2b + 3d = –1

2a – 4c = –4
b – 2d = 0

Note that we have omitted the 3 equations obtained by equating elements of the last
columns of these matrices because the information so obtained would be just a repeat of
that gained by equating elements of the second columns. The augmented matrix of the
above system is

 2 0 8 0 8
 
 0 1 0 4 6 

 −4 0 6 0 6 
 
0 −2 0 3 −1 

 2 0 −4 0 −4 
 
 0 1 0 −2 0 

Supplementary Exercises 1 55

The reduced row-echelon form of this matrix is

1 0 0 0 0
 
 0 1 0 0 2 

0 0 1 0 1
 
 0 0 0 1 1 

0 0 0 0 0

 0 0 0 0 0 

Thus a = 0, b = 2, c = 1, and d = 1.

11. The matrix X in Part (a) must be 2 × 3 for the operations to make sense. The matrices in
Parts (b) and (c) must be 2 × 2.

x y
(b) Let X =  . Then
 z w 

X 1 −1 2  =  x + 3y −x 2x + y 
 0 1   z + 3w −z 
 3   2z + w 

If we equate matrix entries, this gives us the equations

x + 3y = –5 x + 3w = 6

– x = –1 – z = –3

2x + y = 0 2z + w = 7

Thus x = 1 and z = 3, so that the top two equations give y = –2 and w = 1. Since
these values are consistent with the bottom two equations, we have that

X = 1 −2 
 
 3 1 

x y
 ,
11. (c) As above, let X =  z w  so that the matrix equation becomes

 3x + z 3y + w   x + 2y 4x   2 −2 
 − = 
 − x + 2z −y + 2w   z + 2w 4z   5 4 

56 Supplementary Exercises 1

This yields the system of equations

2x – 2y + z =2

–4x + 3y + w = –2

–x + z – 2w = 5

–y – 4z + 2w = 4

with matrix

 2 −2 1 0 2 
 
 −4 3 0 1 −2 

 −1 0 1 −2 5 
 
0 −1 −4 2 4

which reduces to

 1 0 0 0 −113 37 
 
 0 1 0 0 −160 37 

 0 0 1 0 −20 37 
 
0 0 0 1 −46 37 

Hence, x = –113/37, y = –160/37, z = –20/37, and w = –46/37.

15. Since the coordinates of the given points must satisfy the polynomial, we have
p(1) = 2 ⇒ a + b + c = 2

p(–1) = 6 ⇒ a – b + c = 6
p(2) = 3 ⇒ 4a + 2b + c = 3

The reduced row-echelon form of the augmented matrix of this system of equations is

1 0 0 1
 
 0 1 0 −2 

 0 0 1 3 

Thus, a = 1, b = – 2, and c = 3.

Supplementary Exercises 1 57

17. We must show that (I – Jn ) (I – 1 Jn ) = I or that (I – 1 Jn ) (I – Jn) = I. (By virtue of
n–1 n–1
Theorem 1.6.3, we need only demonstrate one of these equalities.) We have

( )I − Jn  I − 1 = I2 − 1 1 IJ n − Jn I + 1 1 J n2
 n − 1 Jn  n− n−

= I− n 1 J n + 1 1 J n2
n− n−

But J 2 = n Jn (think about actually squaring Jn ), so that the right-hand side of the above
n

equation is just I, as desired.

19. First suppose that AB–1 = B–1 A. Note that all matrices must be square and of the same
size. Therefore

(AB–1)B = (B–1 A)B

or
A = B–1 AB

so that

BA = B(B–1 AB) = (BB–1)(AB) = AB

It remains to show that if AB = BA then AB–1 = B–1 A. An argument similar to the one given
above will serve, and we leave the details to you.

21. (b) Let the ijth entry of A be aij. Then tr(A) = a11 + a22 + … + ann, so that
tr(kA) = ka11 + ka22 + … + kann
= k (a11 + a22 + … + ann)
= ktr(A)

(d) Let the ijth entries of A and B be aij and bij, respectively. Then
tr(AB) = a11b11 + a12b21 + … + a1nbn1
+ a21b12 + a22b22 + … + a2nbn2
+…

+ an1b1n + an2b2n + … + annbnn

58 Supplementary Exercises 1

and
tr(BA) = b11a11 + b12a21 + … + b1nan1
+ b21a12 + b22a22 + … + b2nan2
+…
+ bn1a1n + bn2a2n + … + bnnann

If we rewrite each of the terms bijaji in the above expression as ajibij and list the terms
in the order indicated by the arrows below,

tr(BA) = a11b11 + a21b12 + … + an1b1n
+ a12b21 + a22b22 + … + an2b2n
+…
+ a1nbn1 + a2nbn2 + … + annbnn

then we have tr(AB) = tr(BA).

25. Suppose that A is a square matrix whose entries are differentiable functions of x. Suppose
also that A has an inverse, A–1. Then we shall show that A–1 also has entries which are
differentiable functions of x and that

d—A—–1 = –A–1 d—A A–1
dx dx

Since we can find A–1 by the method used in Chapter 1, its entries are functions of x which
are obtained from the entries of A by using only addition together with multiplication and
division by constants or entries of A. Since sums, products, and quotients of differentiable
functions are differentiable wherever they are defined, the resulting entries in the inverse
will be differentiable functions except, perhaps, for values of x where their denominators
are zero. (Note that we never have to divide by a function which is identically zero.) That
is, the entries of A–1 are differentiable wherever they are defined. But since we are assuming
that A–1 is defined, its entries must be differentiable. Moreover,

d ( AA−1) = d ( I ) = 0
dx dx
or

dA A−1 + dA−1 = 0
A
dx dx

Supplementary Exercises 1 59

Therefore dA –1 = dA A –1
A
so that dx dx

dA–1 = – A–1 dA A–1
dx dx

27. (b) Let H be a Householder matrix, so that H = I – 2PPT where P is an n × 1 matrix. Then
using Theorem 1.4.9,
HT = (I – 2PPT)T
= IT – (2PPT)T
= I – 2(PT)T PT
= I – 2 PPT

=H

and (using Theorem 1.4.1) (by the above result)
HT H = H2 (because PT P = I)
= (I – 2PPT)2
= I2 – 2PPT – 2PPT + (–2PPT )2
= I – 4PPT + 4PPT PPT
= I – 4PPT + 4PPT
=I

29. (b) A bit of experimenting and an application of Part (a) indicates that

 an 0 0 

An =  0 bn 0 
 
 d 0 cn 

60 Supplementary Exercises 1

where

d = an–1 + an–2 c + … + acn–2 + cn–1 = an – cn if a ≠ c
a–c

If a = c, then d = nan–1. We prove this by induction. Observe that the result holds
when n = 1. Suppose that it holds when n = N. Then

 aN 0 0   aN +1 0 0
  
AN +1 = AAN = A  0 bN 0  =  0 bN +1 0 
   
 d 0 c N   aN + cd 0 cN +1 

Here

 aN + aN − cN = aN +1 − aN c + aN c − cN +1 = aN +1 − cN +1 if a ≠ c
 c a−c a−c if a = c
aN +  a−c
cd =
( ) ( )


aN + a Na N −1 = N +1 aN

Thus the result holds when n = N + 1 and so must hold for all values of n.

EXERCISE SET 2.1

1. (a) M11 = 7 • 4–(–1) • 1 = 29, M12 = 21, M13 = 27, M21 = –11, M22 = 13, M23 = –5, M31 = –19,
M32 = –19, M33 = 19

(b) C11 = 29, C12 = –21, C13 = 27, C21 = 11, C22 = 13, C23 = 5, C31 = –19, C32 = 19, C33 = 19

3. (a)

A =1⋅ 7 −1 + 2⋅ 6 −1 + 3⋅ 67 = 29 + 42 + 81 = 152

14 −3 4 −3 1

(b) |A| = 1 • M11 – 6 • M21 – 3 • M23 = 152
(c) |A = 6 • M21 + 7 • M22 + 1 • M23 = 152
(d) |A| = 2 • M12 + 7 • M22 + 1 • M32 = 152
(e) |A| = –3 • M31 – 1 • M32 + 4 • M33 = 152
(f) |A| = 3 • M13 + 1 • M23 + 4 • M33 = 152
5. Second column:

A =5⋅ −3 7 = 5 ⋅ −8 = −40

−1 5

61

62 Exercise Set 2.1

7. First column:

k k2 k k2 k k2
A =1⋅ −1⋅ +1⋅ = 0
k k2 k k2 k k2

A = −(k − 1) ⋅ 24 + (k − 3) ⋅ k+1 5 − (k + 1) ⋅ k+1 7

5k 7k 24

9. Third column:

33 5 33 5

A = −3 ⋅ 2 2 −2 − 3 ⋅ 2 2 −2 = −240

2 10 2 41 0

11.

 −3 5 5   3 −5 −5 

adj(A) =  3 −4 −5  ; A = −1; A−1 =  −3 4 5 
 −2 2 3   2 −2 −3 

13.

 2 6 4  1/2 3/2 1 

adj(A) =  0 4 6  ; A = 4; A−1 =  0 1 3/2 
 0 0 2   0 0 1/ 2 

15. (a)

 −4 3 0 −1 

 2 −1 0 0 
A−1 =  
 −7 0 −1 8
 6 0 1 −7 

(b) Same as (a).

(c) Gaussian elimination is significantly more efficient for finding inverses.

Exercise Set 2.1 63

17.

 4 5 0  2

A =  11 1 2  , b =  3  ; A = −132
 1 5 2   1 

|A1| = –36, |A2| = –24, |A3| = –12
x1 = –36/–132 = 3/11, x2 = –24/–132 = 2/11, x3 = 12/–132 = –1/11
19.

 1 −3 1   4 

A=  2 −1 0  , b =  −2  ; A = −11
 4 0 −3   0 

|A1| = 30, |A2| = 38, |A3| = –40
x1 = 30/–11 = –30/11, x2 = 38/–11 = –38/11, x3 = 40/–11 = –40/11
21.

 3 −1 1   4

A =  −1 7 2  , b =  1  ; A =0
 2 6 −1   5 

The method is not applicable to this problem because the determinant of the coefficient
matrix is zero.

23.

 4 1 1 1  6

 3 7 −1 1  , b =  1  ; A = −424
A=   
 7 3 −5 8   −3 
 1 1 1 2  3

64 Exercise Set 2.1

 4 6 1 1

A2  3 1 −1 1  A2 =0
= 7 −3 −5 ;
8

 1 3 1 2 

y = 0/–424 = 0

25. This follows from Theorem 2.1.2 and the fact that the cofactors of A are integers if A has
only integer entries.

27. Let A be an upper (not lower) triangular matrix. Consider AX = I; the solution X of this

equation is the inverse of A. To solve for column 1 of X, we could use Cramer’s Rule. Note

that if we do so then A2, . . ., An are each upper triangular matrices with a zero on the main
diagonal; hence their determinants are all zero, and so x2,1, . . ., xn,1 are all zero. In a similar
way, when solving for column 2 of X we find that x3,2, . . ., xn,2 are all zero, and so on. Hence,
X is upper triangular; the inverse of an invertible upper triangular matrix is itself upper

triangular. Now apply Theorem 1.4.10 to obtain the corresponding result for lower triangular

matrices.

29. Expanding the determinant gives x(b1 – b2) – y(a1 – a2) + a1b2 – a2b1 = 0

x(b1 − b2 ) − y(a1 − a2 ) + a1b1 − a2b1 = 0

y = b1 − b2 x + a1b2 − a2b1
a1 − a2 a1 − a2

which is the slope-intercept form of the line through these two points, assuming that a1 ≠ a2.

31. (a) |A| = A11| • |A22| = (2 • 3 – 4 • –1) • (1 • 2 – 3 • –10 + • –28) = –1080

(b.) Expand along the first column; |A| = –1080.

33. From I4 we see that such a matrix can have at least 12 zero entries (i.e., 4 nonzero entries).
If a 4 × 4 matrix has only 3 nonzero entries, some row has only zero entries. Expanding
along that row shows that its determinant is necessarily zero.

35. (a) True (see the proof of Theorem 2.1.2).

(b) False (requires an invertible, and hence in particular square, coefficient matrix).

(c) True (Theorem 2.1.2).

(d) True (a row of all zeroes will appear in every minor’s submatrix).

EXERCISE SET 2.2

1. (b) We have

2 −1 3 0 −5 − 5 Add –2 times Row 2 to
det( A) = 1 2 4=1 2 4 Row 1 and –5 times
6 0 −13 −14 Row 2 to Row 3.
5 −3
Factor –5 from Row 1
124 and interchange
= (−1)(−5) 0 1 1 Row 1 and Row 2.

0 −13 −14 Add 13 times
Row 2 to Row 3.
1 24
= (−1)(−5) 0 11 By Theorem 2.2.2.
0 −1
0 Add 2 times Row 2
to Row 1 and 3 times
= (−1)(−5)(−1) = −5 Row 2 to Row 3.

2 15 0 5 −1 Add –2 times Row 1 to
det( AT ) = −1 Row 3, and interchange
2 −3 = −1 2 −3 Row 1 and Row 2.
3
46 0 10 −3 By Theorem 2.2.2.

−1 2 −3
= (−1) 0 5 −1

0 0 −1

= (−1)(−1)(5)(−1) = −5

3. (b) Since this matrix is just I4 with Row 2 and Row 3 interchanged, its determinant is –1.

65

66 Exercise Set 2.2

5.

031 112 Interchange
Row 1 and
det( A) = 1 1 2 = ( −1 ) 0 3 1 Row 2.

324 324

112 Add –3 times Row 1
= (−1) 0 3 1 to Row 3.

0 −1 −2

11 2 Factor 3
= (−1)(3) 0 1 1 3 from Row 2.

0 −1 −2

= −3 1 12 Add Row 2
0 1 13 to Row 3.
0 0 −5 3

If we factor –5/3 from Row 3 and apply Theorem 2.2.2 we fInd that
det(A) = –3(–5/3)(1) = 5

7.

3 −6 9 1 −2 3 Factor 3 from
4 Row 1 and Add
det( A) = −2 7 −2 = 3 0 3 5 twice Row 1
to Row 2.
015 01

1 −2 3 Factor 3 from
= (3)(3) 0 1 43 Row 2 and
0 11 3 subtract Row 2
0 from Row 3.

 11  1 −2 3 Factor 11/3
 3  from Row 3.
= 9 0 1 43

0 01

= 9(11 3)(1) = 33

Exercise Set 2.2 67

9. Interchange
Row 1 and
2131 1011 Row 2.

1 0 1 1 = ( −1 ) 2 1 3 1 Add –2 times Row 1
det( A ) = to Row 2.
0210 0210
Add –2 times Row 2
0123 0123 to Row 3; subtract
Row 2 from Row 4.
1 0 11
0 1 1 −1 Add Row 3 to Row 4.
= (−1) 2 10
0 1 23 Add 2 times Row 1 to
0 Row 2; add –2 times
Row 3 to Row 4.
1 011
0 1 1 −1 Add –1 times
= (−1) 0 −1 2 Row 4 to Row 5.
0 014
0

1 011
0 1 1 −1
= (−1) 0 −1 2
0 006
0

= (−1)(−1)(6)(1) = 6

11. 1 3 1 5 3
−2 −7 0 −4 2

det( A ) = 0 0 1 0 1
0 02 1 1
0 00 1 1

13153
0 −1 2 6 8
=0 0 1 0 1
0 0 0 1 −1
00011

13153
0 −1 2 6 8
=0 0 1 0 1
0 0 0 1 −1
00002

Hence, det(A) = (–1)(2)(1) = –2.

68 Exercise Set 2.2

13.

11 1
det( A) = a b c

a2 b2 c2

1 1 1 Add –a times Row 1 to
=0 b−a c−a Row 2; add –a2 times
b2 − a2 c2 − a2 Row 1 to Row 3.
0

Since b2 – a2 = (b – a)(b + a), we add –(b + a) times Row 2 to Row 3 to obtain

1 11
det( A) = 0
b−a c−a
0
( )0 c2 − a2 − ( c − a )( b + a )

= (b –a)[(c2 – a2) – (c – a)(b + a)]
= (b – a)(c –a)[(c + a) – (b + a)]
= (b – a)(c – a)(c – b)

15. In each case, d will denote the determinant on the left and, as usual, det(A) =

∑ ∑±a1j1a2j2 a3j3, where denotes the sum of all such elementary products.

∑ ∑(a) d =
± (ka1j1)a2j2a3j3 = k ± a1j1 a2j2a3j3 = k det(A)

∑ ∑(b) d = ± a1j2 a2j1 a3j3
± a2j1 a1j2 a3j3 =

Exercise Set 2.2 69

a11 + ka21 a12 + ka22 a13 + ka23
a21 a22 a23
a31 a32 a33

= (a11 + ka21)(a22)(a33) + (a12 + ka22)(a23)(a31)
+ (a13 + ka23)(a21)(a32) – (a13 + ka23)(a22)(a31)
– (a12 + ka22)(a21)(a33) – (a11 + ka21)(a23)(a32)
= a11 a22 a33 + a12 a23 a31 + a13 a21 a32 – a13 a22 a31
– a12 a22 a33 – a11 a23 a32
+ ka21 a22 a33 + ka22 a23 a31 + ka23 a21 a32
– ka23 a22 a31 – ka22 a21 a33 – ka21 a23 a32

a11 a12 a13
= a21 a22 a23

a31 a32 a33

17. (8)

1 −2 3 1 1 −2 31
01
5 −9 6 3 = 3 −5 00
0 −1
−1 2 −6 −2 1 −2

2 8 6 1 0 12

R2 → R2 – 2R1
R3 → R3 + 2R1
R4 → R4 – 2R1

3 −5 1 3 −5 1 −2 = 39
1
= 3 1 −2 0 =3 1 −2 0 =3
37
0 12 −1 3 70

R3 → R3 + R1

69

70 Exercise Set 2.2

(9)

2131 0 1 1 −1 1 1 −1

1011 10 1 1 = ( −1 ) 2 1 0
=
0210 02 1 0
12 3
0123 01 2 3

R1 → R1 – 2R2 R3 → R3 + 3R1

1 1 −1 1
= (−1) 2 0 = ( −1 )( −1 ) 2 =6
1 45
4
50

(10) 0 111 −1 0 −1 0

1 1 1 1 1 1 1 1
2 2 2 2 2 2
=
2 11 0 2 11 0
3 33 3 33

− 13 2 0 0 − 13 2 0 0
3 3

R1 → R1 – 2R2

−1 0 −1 1 10 1 1
2
=1 2 1 ( ) ( )1=12 1 1 = 1 − 1 − 1
2 3 3 3 2 3 3 3 2 3 3 3

− 1 2 0 − 1 2 0
3 3 3 3

R1 → R1 + 3R2

= −1  2 + 1 = −1
6  3 3 6

Exercise Set 2.2 71

(11)

13 1 5 3 1 3 1 53 −1 2 6 8
−2 −7
0 −4 2 0 −1 2 6 8 0101
00
00 1 0 1=0 0 1 0 1= 0211
00
2 1 1 0 0 2 11 0011

0 1 1 0 0 0 11

R2 → R2 + 2R1

1 01 101 0 1
= −2
= ( −1 ) 2 1 1 = ( −1 ) 2 0 0 = ( −1 ) ( −2 ) 11

011 011

R2 → R2 – R3

19. Since the given matrix is upper triangular, its determinant is the product of the diagonal
elements. That is, the determinant is x(x + 1)(2x – 1). This product is zero if and only if
x = 0, x = – 1, or x = 1/2.



EXERCISE SET 2.3

1. (a) We have

det( A) = −1 2
= −4 − 6 = −10
34

and

−2 4 = (−2)(8) − (4)(6) = −40 = 22(−10)
det(2 A) =
68

5. (a) By Equation (1),
det(3A) = 33 det(A) = (27)(–7) = –189

(c) Again, by Equation (1), det(2A–1) = 23 det(A–1). By Theorem 2.3.5, we have

det(2A–1) = 8 = –87
det(A)

(d) Again, by Equation (1), det(2A) = 23 det(A) = –56. By Theorem 2.3.5, we have

det[(2A)–1] = 1 = –1
det(2A) 56

73

74 Exercise Set 2.3

(e)

agd adg

b h e =− b e h Interchange Columns
2 and 3.
ci f cfi
Take the transpose of
ab c the matrix.
=− d e f .

gh i

=7

7. If we replace Row 1 by Row 1 plus Row 2, we obtain

b+c c+a b+a a+b+c b+c+a c+b+a

a b c = a b c =0

11 f 11 f

because the first and third rows are proportional.

13. By adding Row 1 to Row 2 and using the identity sin2 x + cos2 x = 1, we see that the
determinant of the given matrix can be written as

sin2 α sin2 β sin2 γ
1 1 1
1 1 1

But this is zero because two of its rows are identical. Therefore the matrix is not invertible.

15. We work with the system from Part (b).
(i) Here

det(λ I − A) = λ − 2 3  = (λ − 2)(λ − 3) − 12 = λ 2 − 5λ − 6
 − 3
 −4 λ

so the characteristic equation is λ2 – 5λ – 6 = 0.

Exercise Set 2.3 75

(ii) The eigenvalues are just the solutions to this equation, or λ = 6 and λ = –1.

(iii) If λ = 6, then the corresponding eigenvectors are the nonzero solutions x =  x1 
to the equation  x2 
 

 6−2 −3   x1  =  4 −3   x1  =  0 
 6−3   x2   −4 3   x2   0 
 −4        

The solution to this system is x1 = (3/4)t, x2 = t, so x =  ( 3 4 )t  is an eigenvector
whenever t ≠ 0.  t 
 

If λ = –1, then the corresponding eigenvectors are the nonzero solutions

 x1  equation
x= x2  to the



 −3 −3   x1  =  0 
   x2   0 
 −4 −4     

If we let x1 = t, then x2 = –t, so x =  t ≠
  is an eigenvector whenever t 0.
 −t 

It is easy to check that these eigenvalues and their corresponding eigenvectors satisfy

the original system of equations by substituting for x1, x2, and λ. The solution is valid for all
values of t.

17. (a) We have, for instance,

a1 + b1 c1 + d1 = a1 + b1 c1 + d1 + a1 + b1 c1 + d1
a2 + b2 c2 + d2 a2 c2 b2 d2

= a1 c1 + b1 d1 + a1 c1 + b1 d1
a2 c2 a2 c2 b2 d2 b2 d2

The answer is clearly not unique.

76 Exercise Set 2.3

19. Let B be an n × n matrix and E be an n × n elementary matrix.
Case 2: Let E be obtained by interchanging two rows of In. Then det(E) = –1 and EA is

just A with (the same) two rows interchanged. By Theorem 2.2.3, det(EA) = –det(A) =
det(E) det(A).

Case 3: Let E be obtained by adding a multiple of one row of In to another. Then det(E)
= 1 and det(EA) = det(A). Hence det(EA) = det(A) = det(E) det(A).

21. If either A or B is singular, then either det(A) or det(B) is zero. Hence, det(AB) = det(A)
det(B) = 0. Thus AB is also singular.

23. (a) False. If det(A) = 0, then A cannot be expressed as the product of elementary
matrices. If it could, then it would be invertible as the product of invertible matrices.

(b) True. The reduced row echelon form of A is the product of A and elementary matrices,
all of which are invertible. Thus for the reduced row echelon form to have a row of
zeros and hence zero determinant, we must also have det(A) = 0.

(c) False. Consider the 2 × 2 identity matrix. In general, reversing the order of the
columns may change the sign of the determinant.

(d) True. Since det(AAT ) = det(A) det(AT ) = [det(A)]2, det(AAT ) cannot be negative.

EXERCISE SET 2.4

1. (a) The number of inversions in (4,1,3,5,2) is 3 + 0 + 1 + 1 = 5.
(d) The number of inversions in (5,4,3,2,1) is 4 + 3 + 2 + 1 = 10.

35
3. = 12 − (−10) = 22

−2 4

5. −5 6
= (−5)(−2) − (−7)(6) = 52
−7 −2

7. a − 3 5 = (a − 3)(a − 2) − (−3)(5) = a2 − 5a + 21
−3 a − 2

−2 1 4
9. 3 5 −7 = (−20 − 7 + 72) − (20 + 84 + 6) = −65

162

300
11. 2 −1 5 = (12 + 0 + 0) − (0 + 135 + 0) = −123

1 9 −4

77

78 Exercise Set 2.4

13. (a)

λ−2 1 = (λ − 2)(λ + 4) + 5
det ( A) =
−5 λ + 4

= λ2 + 2λ – 3 = (λ – 1)(λ + 3)

Hence, det(A) = 0 if and only if λ = 1 or λ = –3.

15. If A is a 4 × 4 matrix, then

∑det(A) = (–1)p a1i1a2i2 a3i3 a4i4

where p = 1 if (i1, i2, i3, i4) is an odd permutation of {1,2,3,4} and p = 2 otherwise. There
are 24 terms in this sum.

17. (a) The only nonzero product in the expansion of the determinant is
a15a24a33a42a51 = (–3)(–4)(–1)(2)(5) = –120

Since (5,4,3,2,1) is even, det(A) = –120.
(b) The only nonzero product in the expansion of the determinant is

a11a25a33a44a52 = (5)(–4)(3)(1)(–2) = 120
Since (1,5,3,4,2) is odd, det(A) = –120.

19. The value of the determinant is
sin2 θ – (–cos2 θ) = sin2 θ + cos2 θ = 1

The identity sin2 θ + cos2 θ = 1 holds for all values of θ.

21. Since the product of integers is always an integer, each elementary product is an integer.
The result then follows from the fact that the sum of integers is always an integer.

23. (a) Since each elementary product in the expansion of the determinant contains a factor
from each row, each elementary product must contain a factor from the row of zeros.
Thus, each signed elementary product is zero and det(A) = 0.

Exercise Set 2.4 79

25. Let U = [aij] be an n by n upper triangular matrix. That is, suppose that aij = 0 whenever
i > j. Now consider any elementary product a1j1a2j2 … anjn. If k > jk for any factor akjk in this
product, then the product will be zero. But if k ≤ jk for all k = 1, 2, …, n, then k = jk for all
k because j1, j2, …, jn is just a permutation of the integers 1, 2, …, n. Hence, a11a22 … ann
is the only elementary product which is not guaranteed to be zero. Since the column indices

in this product are in natural order, the product appears with a plus sign. Thus, the

determinant of U is the product of its diagonal elements. A similar argument works for

lower triangular matrices.



SUPPLEMENTARY EXERCISES 2

1.

x −4
5

y 3 3 + 4 3 x+ 4 y
x′ = 5 x y
= =
5 5

3 −4 9 + 16 5 5
55 25 25

43

55

3
x

5

4 y 3y − 4 −4 x+ 3y
5 5 x
y′ = = =
5

3 −4 1 5 5
55

43

55

3. The determinant of the coefficient matrix is

1 1α 11 α 11

1 1 β = 0 0 β − α = –(β – a) αβ = –(β – a)(β – a)

αβ1 αβ 1

The system of equations has a nontrivial solution if and only if this determinant is zero;
that is, if and only if α = β. (See Theorem 2.3.6.)

81

82 Supplementary Exercises 2

5. (a) If the perpendicular from the vertex of angle α to side a meets side a between angles
β and γ, then we have the following picture:

c= b
>
C
a1
a2

a

Thus cos β = a1 and cos γ = a2 and hence
cb
a = a1 + a2 = c cos β + b cos γ

This is the first equation which you are asked to derive. If the perpendicular intersects

side a outside of the triangle, the argument must be modified slightly, but the same
result holds. Since there is nothing sacred about starting at angle α, the same
argument starting at angles β and γ will yield the second and third equations.

Cramer’s Rule applied to this system of equations yields the following results:

(b.) acb

b0a

cos α = c a 0 = a(–a2 + b2 + c2 ) = b2 + c2 − a2
0cb 2abc 2bc

c0a

ba0

cos β = 0ab = b(a2 − b2 + c2 ) = a2 + c2 − b2
cba 2abc 2ac
bca

2abc

cos γ = 0ca = c(a2 + b2 − c2 ) = a2 + b2 − c2
c0b 2abc 2ab
ba c

2abc

Supplementary Exercises 2 83

7. If A is invertible, then A–1 = 1 adj(A), or adj(A) = [det(A)]A–1. Thus
det (A)
adj(A) = A = I
det (A)

That is, adj(A) is invertible and
[adj(A)]–1 = 1 A
det (A)

It remains only to prove that A = det(A)adj(A–1). This follows from Theorem 2.4.2 and
Theorem 2.3.5 as shown:

A = [ A–1]–1 = 1 adj ( A–1 ) = det (A)adj (A–1)
(A–1 )
det

9. We simply expand W. That is,

dW = d f1( x) f2( x)
dx dx g1( x) g2( x)

= d (f1(x)g2(x) – f2(x)g1(x))
dx

= f1′(x)g2(x) + f1(x)g2′(x) – f2′(x)g1(x) – f2(x)g1′(x)

= [f1′(x)g2(x) – f2′(x)g1(x)] + [f1(x)g2′(x) – f2(x)g1′(x)]

= f1′( x) f2′( x) + f1( x) f2( x)
g1( x) g2( x) g1′( x) g2′ ( x)

84 Supplementary Exercises 2

11. Let A be an n × n matrix for which the entries in each row add up to zero and let x be the
n × 1 matrix each of whose entries is one. Then all of the entries in the n × 1 matrix
Ax are zero since each of its entries is the sum of the entries of one of the rows of A.

That is, the homogeneous system of linear equations

0
Ӈ 
Ax = 

0

has a nontrivial solution. Hence det(A) = 0. (See Theorem 2.3.6.)

13. (a) If we interchange the ith and jth rows of A, then we claim that we must interchange the
ith and jth columns of A–1. To see this, let

 Row 1 
 
A =  Row 2  and A−1=  Col. 1, Col. 2, ⋅⋅⋅ , Col. n
 Ӈ
 
 
Row n 

where AA–1 = I. Thus, the sum of the products of corresponding entries from Row s in
A and from Column r in A–1 must be 0 unless s = r, in which case it is 1. That is, if
Rows i and j are interchanged in A, then Columns i and j must be interchanged in A–1
in order to insure that only 1’s will appear on the diagonal of the product AA–1.

(b) If we multiply the ith row of A by a nonzero scalar c, then we must divide the ith
column of A–1 by c. This will insure that the sum of the products of corresponding
entries from the ith row of A and the ith column of A–1 will remain equal to 1.

(c) Suppose we add c times the ith row of A to the jth row of A. Call that matrix B. Now
suppose that we add –c times the jth column of A–1 to the ith column of A–1. Call that
matrix C. We claim that C = B –1. To see that this is so, consider what happens when

Row j → Row j + c Row i [in A]

Column i → Column i – c Column j [in A–1]

The sum of the products of corresponding entries from the jth row of B and any kth
column of C will clearly be 0 unless k = i or k = j. If k = i, then the result will be c –
c = 0. If k = j, then the result will be 1. The sum of the products of corresponding
entries from any other row of B—say the rth row—and any column of C—say the kth
column—will be 1 if r = k and 0 otherwise. This follows because there have been no
changes unless k = i. In case k = i, the result is easily checked.

Supplementary Exercises 2 85

15. (a) We have

det(λ I − A) = λ − a11 −a12 −a13
−a21 λ − a22 −a23
−a31 −a32 λ − a33

If we calculate this determinant by any method, we find that

det(λI – A) = (λ – a11)(λ – a22)(λ – a33) – a23a32 (λ – a11)
–a13a31(λ – a22) – a12a21(λ – a33)
–a13a21a32 – a12a23a31

= λ3 + (–a11 – a22 – a33)λ2
+ (a11a22 + a11a33 + a22a33 – a12a21 – a13a31 – a23a32)λ
+ (a11a23a32 + a12a21a33 + a13a22a31
–a11a22a33 – a12a23a31 – a13a21a32)

(b) From Part (a) we see that b = –tr(A) and d = –det(A). (It is less obvious that c is the
trace of the matrix of minors of the entries of A; that is, the sum of the minors of the
diagonal entries of A.)

17. If we multiply Column 1 by 104, Column 2 by 103, Column 3 by 102, Column 4 by 10, and
add the results to Column 5, we obtain a new Column 5 whose entries are just the 5
numbers listed in the problem. Since each is divisible by 19, so is the resulting determinant.



TECHNOLOGY EXERCISES 2

T3. Let y = ax3 + bx3 + cx + d be the polynomial of degree three to pass through the four
given points. Substitution of the x and y coordinates of these points into the equation of the
polynomial yields the system
7 = 27a + 9b + 3c + d
–1 = 8a + 4b + 2c + d
–1 = a + b + c + d
1 = 0a + 0b + 0c + d

Using Cramer’s Rule,

7 9 31 27 73 1
−1 2 1
−1 4 2 1 8 −1 1 1
1 = −24 = −2
−1 1 1 1 1 10
12 12
a = 1 0 0 1 = 12 = 1, b = 0
27 9 3 1 12

8 4 21

1 111

0 001

27 97 1 27 93 7
8 4 −1 1 8 42 −1
1 1 −1 1 1 11 −1
0 01 1 = −12 = −1, 0 00
c= d= 1 = 12 = 1
12 12 12 12

Plot. y = x3 – 2x2 – x + 1

87

88 Technology Exercises 2

(3, 7)

(0, 1)

(1, −1) (2, −1)

EXERCISE SET 3.1

1. (a) z (c) z
x (3, 4, 5) (3, 4, 5) y
y
x

(e) ( 3, 4, 5) z (j) z
(3, 0, 3)
y x y
x

→

→3. (a) P1P2 = (3 – 4, 7 – 8) = (–1, –1)
→
→→(e) P1P2 = (–2 – 3, 5 + 7, –4 –2) = (–5, 12, –6)

5. (a) Let P = (x, y, z) be the initial point of the desired vector and assume that this vector

→

has the same length as v. Since PQ has the same direction as v = (4, –2, –1), we have

the equation

→

PQ = (3 – x, 0 – y, –5 – z) = (4, –2, –1)

89

90 Exercise Set 3.1

If we equate components in the above equation, we obtain
x = –1, y = 2, and z = –4

→

Thus, we have found a vector PQ which satisfies the given conditions. Any positive

→

multiple k PQ will also work provided the terminal point remains fixed at Q. Thus, P
could be any point (3 – 4k, 2k, k – 5) where k > 0.
(b) Let P = (x, y, z) be the initial point of the desired vector and assume that this vector

→

has the same length as v. Since PQ is oppositely directed to v = (4, –2, –1), we
have the equation

→

PQ = (3 – x, 0 – y, –5 –z) = (–4, 2, 1)

If we equate components in the above equation, we obtain
x = 7, y = –2, and z = –6

→

Thus, we have found a vector PQ which satisfies the given conditions. Any positive

→

multiple k PQ will also work, provided the terminal point remains fixed at Q. Thus,
P could be any point (3 + 4k, –2k, –k – 5) where k > 0.

7. Let x = (x1, x2, x3). Then
2u – v + x = (–6, 2, 4) – (4, 0, –8) + (x1, x2, x3)
= (–10 + x1, 2 + x2, 12 + x3)

On the other hand,

7x + w = 7(x1, x2, x3) + (6, –1, –4)
= (7x1 + 6, 7x2 – 1, 7 x3 – 4)

If we equate the components of these two vectors, we obtain

7x1 + 6 = x1 – 10
7x2 – 1 = x2 + 2
7x3 – 4 = x3 + 12

Hence, x = (–8/3, 1/2, 8/3).

Exercise Set 3.1 91

9. Suppose there are scalars c1, c2, and c3 which satisfy the given equation. If we equate
components on both sides, we obtain the following system of equations:

–2c1 – 3c2 + c3 = 0
9c1 + 2c2 + 7c3 = 5

6c1 + c2 + 5c3 = 4
The augmented matrix of this system of equations can be reduced to

 2 3 −1 0 
 
 0 2 −2 −1 

 0 0 0 −1 

The third row of the above matrix implies that 0c1 + 0c2 + 0c3 = –1. Clearly, there do not
exist scalars c1, c2, and c3 which satisfy the above equation, and hence the system is
inconsistent.

11. We work in the plane determined by the three points O = (0, 0, 0), P = (2, 3, –2), and Q =

→

(7, –4, 1). Let X be a point on the line through P and Q and let t PQ (where t is a positive,

→real number) be the vector with initial point P and terminal point X. Note that the length
→ →
of t PQ is t times the length of PQ . Referring to the figure below, we see that

OP→+ t → = →

PQ OX

and

OP→+ → = →

PQ OQ

tPQ Q
X

P

O

92 Exercise Set 3.1

Therefore,

→ = OP→+ → →

OX t( OQ – OP )

= (1 – t) OP→+ t →

OQ

(a) To obtain the midpoint of the line segment connecting P and Q, we set t =1/2. This
gives

→ = 1 OP→+ 1 →

OX 22 OQ

= 1 (2,3,−2)+ 1 (7,−4,1)
22

=  9 ,− 1 ,− 1 
 
2 2 2

(b) Now set t = 3/4.This gives

→ = 1 (2, 3, −2) + 3 (7, −4,1) =  23 , − 9 , 1 
 
OX 4 4  4 4 4

13. Q = (7, –3, –19)

17. The vector u has terminal point Q which is the midpoint of the line segment connecting P1
and P2.

1 (OP2 − OP1)
2

Q P2

P1

OP1 OP2 −OP1

O OP2 − OP1

Exercise Set 3.1 93

19. Geometrically, given 4 nonzero vectors attach the “tail” of one to the “head” of another and
continue until all 4 have been strung together.The vector from the “tail” of the first vector
to the “head” of the last one will be their sum.

y
z

w
x

x+y+z+w